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[2020-02-06 Thu 13:30]
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Session 57: Work and Line Integrals | Part B: Vector Fields and Line Integrals | 3. Double Integrals and Line Integrals in the Plane | Multivariable Calculus | Mathematics | MIT OpenCourseWare

1 Chalkboard

Figure 1: Work and line integrals

Figure 1: Work and line integrals

Figure 2: Work as integral

Figure 2: Work as integral

Figure 3: Example of work

Figure 3: Example of work

Figure 4: Example cont

Figure 4: Example cont

Figure 5: Another way of calculate line integrals

Figure 5: Another way of calculate line integrals

Figure 6: Compute line integrals in terms of one variable

Figure 6: Compute line integrals in terms of one variable

Figure 7: Example

Figure 7: Example

Figure 8: Not practical example

Figure 8: Not practical example

2 How is a line integrals?

2.1 Front

How is a line integrals?

What do you need and notation

2.2 Back

  • Field: \(\vb{F} = M \vu{i} + N \vu{j} = \ev{M,N}\)
  • Curve: \(\vb{r}(t) = x(t)\vu{i} + y(t)\vu{j} = \ev{x,y} \implies \dd{\vb{r}} = \ev{\dd{x}, \dd{y}}\)

Line integral

\({\displaystyle \int_C \vb{F} \cdot \dd{\vb{r}} = \int_C \ev{M,N} \cdot \ev{\dd{x},\dd{y}} = \int_C M \dd{x} + N \dd{y}}\)

3 How line integrals arise?

3.1 Front

How line integrals arise?

A force \(\vb{F}\) being applied over a displacement \(\vb{\Delta r}\)

Work as line integral

3.2 Back

Work is force times distance, but only the component of the force in the direction of the displacement does any work.

Work = \(\abs{\vb{F}} \cos \theta \abs{\vb{\Delta r}} = \vb{F} \cdot \vb{\Delta r}\)

For a variable force applied over a curve the total work is found by “summing” the infinitesimal pieces

Total work is \({\displaystyle \lim_{n \to \infty} \sum_{i=1}^n \vb{F_i} \cdot \Delta \vb{r}}\)

Line integral: \({\displaystyle \int_C \vb{F} \cdot \dd{\vb{r}}}\)

4 How can we compute a line integral?

4.1 Front

How can we compute a line integral?

Using the example: \({\displaystyle I = \int_C x^2 y \dd{x} + (x-2y)\dd{y}}\), over the part of parabola \(y=x^2\) from \((0,0)\) to \((1,1)\)

4.2 Back

  1. Parametrize the curve
    • \(x = t\), \(y = t^2\), \(0 \geq t \geq 1\)
    • \(\dd{x} = \dd{t}\), \(\dd{y} = 2t \dd{t}\)
  2. Substitute everything in the integral
    • \({\displaystyle \int_0^1 t^2 (t^2) \dd{t} + (t - 2t^2) 2t \dd{t} = \int_0^1 t^4 + 2t^2 - 4t^3 \dd{t} = - \frac{2}{15}}\)

5 Is the line integral dependent of the parametrization?

5.1 Front

Is the line integral dependent of the parametrization?

Using the same integral, but different parametrization of \(C\), is the line integral different?

5.2 Back

No, the line integral is independent of the parametrization

6 What happens if we reverse the direction of the curve in the line integral?

6.1 Front

What happens if we reverse the direction of the curve in the line integral?

\({\displaystyle \int_{-C} \vb{F} \cdot \dd{\vb{r}}}\)

6.2 Back

The line integral changes sign

\({\displaystyle - \int_C \vb{F} \cdot \dd{\vb{r}}}\)

7 Which notation can we use for closed curve on line integral?

7.1 Front

Which notation can we use for closed curve on line integral?

7.2 Back

\({\displaystyle \oint_C \vb{F} \cdot \dd{\vb{r}}}\)